I want to solve a differential equation of second order with complex initial conditions.
u''[τ[t]] = -f[t]u[τ[t]]
u[27] == Exp[-I*k55*τ[27]]/Sqrt[2*k55]
u'[27] == -I*k55*Exp[-I*k55*τ[27]]/Sqrt[2*k55]
Although I have written all the code in detail, I am still facing some problems I might want to review here. The code is as follows:
sol = NDSolve[{x''[t] + 3 x'[t] Sqrt[(x'[t])^2/6 + (1 - Exp[-x[t] Sqrt[2/3]])^2/4] +
Sqrt[3/2] Exp[-x[t] Sqrt[2/3]] (1 - Exp[-x[t] Sqrt[2/3]]) == 0,
a'[t]/a[t] == Sqrt[(x'[t])^2/6 + (1 - Exp[-x[t] Sqrt[2/3]])^2/4],
τ'[t] == 1/a[t],
x'[0] == -0.008226306418212731,
x[0] == 5.630991866033891,
a[0] == 1,
τ[149.4517772937791] == 0},
{x, τ, a},
{t, 0, 500}]
a = a[t] /. sol;
τ = τ[t] /. sol;
x = x[t] /. sol;
xt = x'[t] /. sol;
att = a''[t] /. sol;
tEnd = t /. FindRoot[(att == 0), {t, 0, 150}]
aEnd = a /. t -> tEnd;
H = Sqrt[(xt)^2/6 + (1 - Exp[-x*Sqrt[2/3]])^2/4];
V = 3/4 (1 - Exp[-Sqrt[2/3] x])^2;
Vt = Sqrt[3/2] Exp[-x Sqrt[2/3]] (1 - Exp[-x Sqrt[2/3]]);
Vtt = -Exp[-2 Sqrt[2/3]*x]*(Exp[Sqrt[2/3]*x] - 2);
t55 = t /. FindRoot[(aEnd/a == E^55), {t, 0, 150}];
τ55 = τ /. t -> t55;
a55 = a /. t -> t55;
eps1 = (1/2) (Vt/V)^2;
eta = Vtt/V;
eps2 = -4 eps1 + 2 eta;
k55 = a55*(H /. t -> t55);
ν = 3/2 + eps1 + 1/2*eps2;
f = k55^2 - (ν^2 - 1/4)/(τ^2);
solu = NDSolve[{u''[t] == -u[t]*f, u[27] == Exp[I*k55*τ[27]]/Sqrt[2*k55], u'[27]
== -I*k55*Exp[I*k55*τ[27]]/(a55*Sqrt[2*k55])}, u[t], {t, 27, 40}]
My problem here is that when trying to get solu (run the NDSolve for u(τ(t))
), it does give a solution but for a minimal interval (t=27 to t=27.015097919331026642336'20) when the desired interval goes from t=27 to t=40.
How do I solve it?
I took the anzats $u(t)=e^{\int v(t)dt}$, therefore, I got that $v'+v^2+f=0$. I got $v$ using
solv = NDSolve[{v'[t] + (v[t])^2 + f == 0, v[0] == -I*k55*Exp[-I*k55*τ[27]]}, v[t], {t, 27, 40}]
However, it has the same problems as the solution for $u$. The solution for either $v$ of $u$ has to go from $t=27$ to $t=40$ approximately. Even when trying to solve the systems with f at first order with FindRoot[f, {t, 36}][[1, 2]]; Series[f, {t, %, 1}] // Normal, it does not give any satisfactory solution to the problem. It highly oscillates.
NDSolveValue::mxst: Maximum number of 1000000 steps reached at the point t == 27.01283361800840725823760157657196543473`20.
Update:
I am using new initial conditions
solu = NDSolve[{u''[t] == -f*u[t],
u[tEnd] == Exp[-I*kEnd*\[Tau]End]/Sqrt[2*kEnd],
u'[tEnd] == -I*kEnd*Exp[-I*kEnd*\[Tau]End]/(aEnd*Sqrt[2*kEnd])},
u, {t, 0, tEnd}, WorkingPrecision -> 20, MaxSteps -> 1000000];
And even though I get a solution, I cannot plot it because, for example, if I do
LogLinearPlot[Re[H/(a*xt) (u[t] /. solu)], {t, 60, tEnd}]
It shows
InterpolatingFunction::dmval: Input value {60.0011} lies outside
the range of data in the interpolating function. Extrapolation
will be used.
It happens with any initial value for $t$ to begin the Plotting, replace 60.0011 with any general number you may think of. Why does it happen?
Edit:
Now I understand it happens because the solution runs backwards, so I only have a solution from t=149.467... to t=tEnd. How can I get the entire solution? I know it seems complicated, but my initial conditions come from $u_k(\tau)=\lim_{k\gg aH}\frac{e^{-ik\tau}}{\sqrt{2k}}$, where $aH=k_{55}$, and when $t\rightarrow\infty$, it gives a stable limit value.
